Question

A theater has a seating capacity of 750 and charges $2 for children, $4 for students, and $6 for adults. At a certain screening with full attendance, there were half as many adults as children and students combined. The receipts totaled $3300. How many children attended the show? (Let x, y, and z denote the number of children, students, and adults, respectively.)

Answer 1

100 children

Explanation:

Let x, y, and z denote the number of children, students, and adults, respectively.

A theater has a seating capacity of 750

So,
`x+y+z= 750` ---A

Theater charges $2 for children, $4 for students, and $6 for adults.

Cost for x children = 2x

Cost for y students = 4y

Cost for y adults = 6z

The receipts totaled $3300

So,
`2x+4y+6z=3300` ---B

Now we are given that there were half as many adults as children and students combined.

So,
`z=(1)/(2)(x+y)` ---C

`2z=x+y`

Substitute the value of x+y in A

`2z+z= 750`

`3z= 750`

`z= 250`

Substitute the value of z in A and B

In A

`x+y+250= 750`

`x+y=500` ---D

In B

`2x+4y+6(250)=3300`

`2x+4y=1800` ---E

Solve D and E

Substitute the value of x from D in E

`2(500-y)+4y=1800`

`1000-2y+4y=1800`

`2y=800`

`y=400`

Substitute the value of y in D

`x+400=500`

x=100

Hence 100 children attended the show